Science:Math Exam Resources/Courses/MATH257/December 2011/Question 04 (a)

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MATH257 December 2011

• Q1 • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) •

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• December 2011 •

Question 04 (a)
Consider the following problem for the heat equation with a time-dependent source term, and mixed boundary conditions: $u_{t}=u_{xx}+t,\quad 0<x<1,\ t>0,$ $u_{x}(0,t)=0,\quad u(1,t)=0,\quad u(x,0)=1.$ Briefly describe how you would use the method of finite differences to find an approximate solution to this problem. Use the notation $u_{n}^{k}\approx u(x_{n},t_{k})$ to denote the values of $u$ on the finite difference mesh, and include how you propose to incorporate the boundary and initial conditions. In case it is useful, the Taylor expansion formula is $f(x+\Delta x)=f(x)+f'(x)\Delta x+{\frac {1}{2}}f''(x)(\Delta x)^{2}+O((\Delta x)^{3}).$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
For $\Delta t$ small, Taylor expansion of $u(x,t+\Delta t)$ with respect to time variable about (x,t) gives $~u(x,t+\Delta t)=u(x,t)+u_{t}(x,t)\Delta t+O((\Delta t)^{2})$ This reminds us of the forward difference approximation for the time derivative: $u_{t}(x,t)\approx {\frac {u(x,t+\Delta t)-u(x,t)}{\Delta t}}$ which has error of $O(\Delta t)$ , i.e. first order accurate. Similarly, for $\Delta x$ small, Taylor expansion of $u(x+\Delta x,t)$ with respect to the space variable about (x,t) gives $u(x+\Delta x,t)=u(x,t)+u_{x}(x,t)\Delta x+{\frac {1}{2!}}u_{xx}(x,t)(\Delta x)^{2}+{\frac {1}{3!}}u_{xxx}(x,t)(\Delta x)^{2}O((\Delta x)^{4})$ and $u(x-\Delta x,t)=u(x,t)-u_{x}(x,t)\Delta x+{\frac {1}{2!}}u_{xx}(x,t)(\Delta x)^{2}-{\frac {1}{3!}}u_{xxx}(x,t)(\Delta x)^{3}+O((\Delta x)^{4})$ This reminds us of the centered difference approximation for the second space derivative: $u_{xx}(x,t)\approx {\frac {u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)}{\Delta x^{2}}}$ which has an error of $O(\Delta x^{2})$ , i.e. second order accurate. Putting these back to the equation gives ${\frac {u(x,t+\Delta t)-u(x,t)}{\Delta t}}={\frac {u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)}{\Delta x^{2}}}\quad +\quad t\quad +\quad O(\Delta t,\Delta x^{2})$ With uniform time step $~\Delta t=t_{max}/K$ and mesh spacing $~\Delta x=1/N$ , let $~x_{n}=n\Delta x$ , where $n=0,1,2,...,N$ ; and $~t_{k}=k\Delta t$ , where $k=0,1,2,...,K$ ; and $u_{n}^{k}=u(x_{n},t_{k})$ . Plug these mesh points onto the above gives a linear system with variables $u_{n}^{k}$ . (Note that some of these values are not unknown, but can be calculated directly using the boundary conditions.) Work out an equation for a fixed n and k, showing $u_{n}^{k+1}$ in terms of values from the previous time step, i.e. $u_{n}^{k},u_{n-1}^{k},u_{n+1}^{k}$ . Then consider what should $u_{n}^{k}$ be when $k=0$ and when $n=0,N$ Be reminded that the left Neumann condition $~u_{x}(0,t)=0$ is tricky in that you need again to use the centered difference formula for first derivative to incorporate that. You might want to add a "fictional" point to do that.

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Solution
By the Taylor series expansion, $u(x,t+\Delta t)=u(x,t)+\Delta t\;u_{t}(x,t)+O((\Delta t)^{2}).\,$ If we isolate $u_{t}(x,t)$ in the above, we get that $u_{t}(x,t)={\frac {u(x,t+\Delta t)-u(x,t)}{\Delta t}}.$ Similarly, by Taylor series expansion, we have $u(x+\Delta x,t)=u(x,t)+\Delta x\;u_{x}(x,t)+{\frac {(\Delta x)^{2}}{2!}}u_{xx}(x,t)+{\frac {(\Delta x)^{3}}{3!}}u_{xxx}(x,t)+O((\Delta x)^{4})$ and $u(x-\Delta x,t)=u(x,t)-\Delta x\;u_{x}(x,t)+{\frac {(\Delta x)^{2}}{2!}}u_{xx}(x,t)-{\frac {(\Delta x)^{3}}{3!}}u_{xxx}(x,t)+O((\Delta x)^{4}).$ If we add the two equations above together, we get that $u(x+\Delta x,t)+u(x-\Delta x,t)=2u(x,t)+(\Delta x)^{2}u_{xx}(x,t)+O((\Delta x)^{4}).\,$ If we isolate $u_{xx}(x,t)$ in the above, we get that $u_{xx}(x,t)={\frac {u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)}{(\Delta x)^{2}}}+O((\Delta x)^{2}).$ From these, we see that we can approximate $u_{t}(x,t)$ by $u_{t}(x,t)\approx {\frac {u(x,t+\Delta t)-u(x,t)}{\Delta t}}$ and $u_{xx}(x,t)$ by $u_{xx}(x,t)\approx {\frac {u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)}{(\Delta x)^{2}}}.$ Hence, we can approximate the equation $u_{t}=u_{x}x+t\,$ by ${\frac {u(x,t+\Delta t)-u(x,t)}{\Delta t}}={\frac {u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)}{(\Delta x)^{2}}}+t.$ If we isolate $u(x,t+\Delta t)$ in the above, we get that $u(x,t+\Delta t)=u(x,t)+{\frac {\Delta t}{(\Delta x)^{2}}}[u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)]+t\Delta t.\;\;\;{\text{(1)}}$ Next we will put $N+1$ mesh points in the interval $[0,1]$ . Let $x_{n}=n\Delta x\;\;{\text{for}}\;\;n=0,1,2,\cdots ,N$ where $\Delta x={\frac {1}{N}}.$ Here, $x_{0}=0$ and $x_{N}=1$ . Similarly, we will let the mesh points in $t$ be $t_{k}=k\Delta t\;\;{\text{for}}\;\;k=0,1,2,\cdots$ for some small time step $\Delta t$ . At $x=x_{n}$ and $t=t_{k}$ , equation (1) becomes $u(x_{n},t_{k+1})=u(x_{n},t_{k})+{\frac {\Delta t}{(\Delta x)^{2}}}[u(x_{n+1},t_{k})-2u(x_{n},t_{k})+u(x_{n-1},t_{k})]+t_{k}\Delta t.$ To get the above, we used $t+\Delta t=t_{k+1},x+\Delta x=x_{n+1}\;\;{\text{and}}\;\;x-\Delta x=x_{n-1}$ . Using the notation $u_{n}^{k}\approx u(x_{n},t_{k})$ , we can write the above as $u_{n}^{k+1}=u_{n}^{k}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{n+1}^{k}-2u_{n}^{k}+u_{n-1}^{k}]+t_{k}\Delta t.\;\;(2)$ If we look at the above equation, we see that for each fixed $k$ , it gives us a scheme to compute $u_{n}^{k+1}$ once $u_{n}^{k}$ is known for $n=0,1,2,\cdots ,N$ . For the initial condition $u(x,0)=0$ , we set $u_{n}^{0}=0\;\;{\text{for}}\;\;n=0,1,2,\cdots ,N.$ For the boundary condition $u(1,t)=0$ , we set $u_{N}^{k}=0\;\;{\text{for}}\;\;k=1,2,3,\cdots .$ Now, for the boundary condition $u_{x}(0,t)=0$ , there is more than one way to handle that. One possible way is to set ${\frac {u(0,t)-u(0-\Delta x,t)}{\Delta x}}=0$ which implies $u(0,t)=u(-\Delta x,t)\,$ If we allow an extra mesh point to the left of $x_{0}$ , the above gives $u_{0}^{k}=u_{-1}^{k}.$ Hence, for $n=0$ , equation (2) can be rewritten as ${\begin{aligned}u_{0}^{k+1}&=u_{0}^{k}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{1}^{k}-2u_{0}^{k}+u_{-1}^{k}]+t_{k}\Delta t\\&=u_{0}^{k}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{1}^{k}-u_{0}^{k}]+t_{k}\Delta t.\end{aligned}}$ Referring to the diagram above, to find an approximate solution to this problem, we first set the initial conditions $u_{n}^{0}=0\;\;{\text{for}}\;\;n=0,1,2,\cdots ,N$ and the boundary condition $u_{N}^{k}=0\;\;{\text{for}}\;\;k=1,2,3,\cdots$ Then, we compute $u_{n}^{1}$ from $u_{n}^{0}$ by $u_{0}^{1}=u_{0}^{0}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{1}^{0}-u_{0}^{0}]+t_{0}\Delta t$ and $u_{n}^{1}=u_{n}^{0}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{n+1}^{0}-2u_{n}^{0}+u_{n-1}^{0}]+t_{0}\Delta t\;\;{\text{for}}\;\;n=1,2,3,\cdots ,N.$ Once we have $u_{n}^{1}$ , we can iterate the process to find $u_{n}^{2},u_{n}^{3},\cdots$